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Let $R$ be a commutative, unital ring. Let $F = R^{(X)}$ and $G = R^{(Y)}$ be the free $R$-modules generated two arbitrary sets $X$ and $Y$ respectively. Assume there is a monomorphism $F \hookrightarrow G$. Does this imply that $|X| \leq |Y|$?

In the case where $X$ and $Y$ are finite, the answer is yes and several proofs can be found here and here. Notice that the solutions given in these threads firmly rely the finiteness condition, which leaves me wondering about the infinite case.

According to this answer, the cardinal of $R^{(X)}$ is $\max \left\{|R|,|X|\right\}$. It follows that if a counterexample exists then the following inequalites must be satisfied: $$\aleph_0 \leq |Y| < |X| \leq |R|$$

In particular, $|R| \geq \aleph_1$.

I feel quite out of my depth when it comes to searching for a counterexample because I don't know any $\aleph_1$-sized rings that are simple enough that it is possible to describe the relevant free modules over them... On the other hand, I really have no idea of how to approach a potential proof.

Many thanks.

user1892304
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1 Answers1

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The infinite case follows from the finite case. Specifically, supposing $Y$ is infinite, for each finite subset $S\subseteq Y$, let $A_S\subseteq X$ be the set of basis elements of $F$ which map to elements of $G$ whose support is contained in $S$. Note that our monomorphism $F\to G$ then restricts to a monomorphism $R^{(A_S)}\to R^S$, and thus $|A_S|\leq |S|$. But every element of $X$ is in $A_S$ for some $S$. Since $Y$ has $|Y|$ different finite subsets, and $A_S$ is finite for each one of them, it follows that $|X|\leq |Y|\cdot\aleph_0=|Y|$.

Eric Wofsey
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